A drug made by a pharmaceutical company comes in tablet form

A drug made by a pharmaceutical company comes in tablet form. Each tablet is branded as containing 95 mg of the particular active chemical. However, variation in manufacturing results in the actual amount of the active chemical in each tablet following a normal distribution with mean 95 mg and standard deviation 2.002 mg. Calculate the percentage of tablets that will contain less than 94 mg of the active chemical. You may find this standard normal table useful. Give your answer as a percentage to 2 decimal places. Percentage = % Suppose samples of 12 randomly selected tablets are taken and the amount of active chemical measured. Calculate the percentage of samples that will have a sample mean of less than 94 mg of the active chemical. Give your answer as a percentage to 2 decimal places. Percentage = % The weekly price fluctuation of Fjord stock has a mean of $2.10 and a standard deviation of $0.38. A sample of 107 weekly price fluctuations for this stock have been measured and the mean has been calculated. Calculate the standard error of the mean Give your answer to 3 decimal places. The number of diners at a restaurant each day is recorded and a daily average is calculated every month (assume 30 days in a month). The number of diners each day has a mean of 142 and a standard deviation of 58, but does not necessarily follow a normal distribution. The probability that a daily average over a given month is greater than is 2.5%. Calculate You may find standard normal table useful. Give your answer to 3 decimal places.

Solution

Q3: A drug made by a phramaceutical company....

a)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    94      
u = mean =    95      
          
s = standard deviation =    2.002      
          
Thus,          
          
z = (x - u) / s =    -0.4995005      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -0.4995005   ) =    0.308713417 = 30.87% [answer]

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b)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    94      
u = mean =    95      
n = sample size =    12      
s = standard deviation =    2.002      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.730320487      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z >   -1.730320487   ) =    0.041786516 = 4.18% [answer]
          

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